James' Homework 1
Comments: 2.2, Good. Also remember to mention that the inverse is also a positive real (i.e, inverse closed). For example, Rpos U {0} (i.e R positive with 0) under addition fails to be a group since even though addition is associative, identity is there, even though an inverse exist (adding -x) -x is not in the set. For a first homework in group theory, you did pretty well. SO much better than many ppl who took Math 109 when I graded it (I even have the grades on coursework). 2.8. Flawless. 3.1i Good. Same comment as 2.2 3.1ii. GOOD! You even checked the inverse condition required in 2.2 and 3.1i. You definitely get it. Also, you get the fact you have to also show that the binary operator is closed. 3.2. Even a bigger GOOD than the previous (limited font) 3.3 Dude, you overthought the question. The n is fixed. For example, say n = 4. Then -1, 1, i, -i are in G. The inverse proof you did is right...but the (a^k *b^m) => (a*b)^{k+m} step is incorrect i^4 = -1, -1^2 = 1, but -i^6 = -1. BUT! If k = m then it becomes so much simpler. Try to this one last part again instead of me spoiling the punchline. 3.10 Easy solution would be to simply cite 3.5 and 3.9. Kinda lost you on the second sentence (how do you know this makes it not a group)? 4.2. Bah, Practice computation. It does have its uses. I see that you get Z_n and the concept of order. 4.4. Prime factors? This is a general group. The prove is like proving 1 to 1 in the bijection in fermat's little theorem (animals in cages). Suppose for x and y distinct, gx was the same as gy. Then, gx = gy. But multiply both sides, on the left, by g inverse. This means x = y. But the definition of distinct means x and y are different, a contradiction. Can you make the same argument for xg? 4.6 I will explain this over the phone, but you cannot take square roots because there is no idea for "positive" and so there is a well defineness issue (There can exist multiple different square roots, how do you know you are choosing the same one). I think you have a right idea, but there is a different trick. Let me thick of it after I get some more starcraft 2. 4.8 SPOT ON! Noticing that (XY)^(N+1)=XY, flawless. Great job and a pretty solution. 5.5. I think I am going to demonstrate this one: I never did teach "iff (if and only_" 5.7. The identity part correct, but the fact that x^-1 exists is because if x^n = e then multiply both sides by x^-1 for n times. Then e = (x^-1)^n. Hm...how do we know that xy has finite order if x and y do? This is nontrivial. WAIT, we assumed the group G is ABELIAN. So x^n = e, x^m = e implies (xy)^mn = xyxyxyxyxyxy...= xxxxxxyyyyy = x^mn*y^mn= (x^n)^m * (x^m)^n = 1^m * 1^n = 1. I deserve a cookie. 6.1 ^ ^ good, everyone needs to do this at least once in their life 6.3 Good. Just because it is a permutation with different names doesn't mean it isnt just S3. 6.11. HAHHAHAHAHAHAH I did the same thing my first time, and then realized I could just apply some theorem. page_1-001.jpg|Jame's Homework 1 p.1 hw1page_2-001.jpg|James' Homework 1 p.2 page_3-001.jpg|James' Homework 1 p.3 page_4-001.jpg|James' Homework 1 p.4 page_5-001.jpg|James' Homework 1 p.5 page_6-001.jpg|James' Homework 1 p.6 page_7-001.jpg|James' Homework 1 p.7 page_8-001.jpg|James' Homework 1 p.8